The length of focal chord AB of ellipse is

Question

The length of focal chord AB of ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ is $(Given A = (\frac{8}{5}, \frac{3 \sqrt{3}}{5}))$

Smart Achievers · Accepted Answer

(2)

$∵$ $2 \cos θ_{1} = \frac{8}{5} \Rightarrow \cos θ_{1} = \frac{4}{5} \Rightarrow \tan \frac{θ_{1}}{2} = \sqrt{\frac{1 - \frac{4}{5}}{1 + \frac{4}{5}}} = \frac{1}{3} and e = \frac{1}{2}$

$∵$ $\tan \frac{θ_{2}}{2} \times \frac{1}{3} = \frac{- \frac{1}{2}}{\frac{3}{2}} \Rightarrow \frac{θ_{2}}{2} = \frac{3 π}{4} \Rightarrow θ_{2} = \frac{3 π}{2}$

$B = (0, - \sqrt{3}) \Rightarrow A B = 2 + 2 - \frac{1}{2} \times \frac{8}{5} = 4 - \frac{4}{5} = \frac{16}{5}$

Solution

Q.
The length of focal chord AB of ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ is $(Given A = (\frac{8}{5}, \frac{3 \sqrt{3}}{5}))$

1 $\frac{4}{5}$

2 $\frac{16}{5}$

3 $\frac{32}{5}$

4 $\frac{64}{5}$

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Solution

Q. The length of focal chord AB of ellipse x24+y23=1 is (Given A=(85,335))

1 45

2 165

3 325

4 645